Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;

public class Main {
	static class Reader
    {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[256]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1)
            {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
            {
                ret = ret * 10 + c - '0';
            }  while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.')
            {
                while ((c = read()) >= '0' && c <= '9')
                {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
	
	
	public static void main(String args[])throws IOException{
		Reader in = new Reader();
		int n = in.nextInt();
		int m = in.nextInt();
		int[][] g = new int[n][m];
		int[] min_r = new int[n];
		int[] min_c = new int[m];
		int[][] h = new int[n][m];
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				g[i][j] = in.nextInt();
				h[i][j]=g[i][j];
			}
		}
		ArrayList<String> arli = new ArrayList<>();
		for(int i=0;i<n;i++){
			int min=Integer.MAX_VALUE;
			for(int j=0;j<m;j++){
				if(g[i][j]<min)
					min=g[i][j];
			}
			min_r[i]=min;
		}
		for(int i=0;i<n;i++){
			int x=min_r[i];
			while(min_r[i]>0){
				arli.add("row "+(i+1));
				min_r[i]--;
			}
			for(int j=0;j<m;j++)
				g[i][j]-=x;
		}

		for(int i=0;i<m;i++){
			int min=Integer.MAX_VALUE;
			for(int j=0;j<n;j++){
				if(g[j][i]<min)
					min=g[j][i];
			}
			min_c[i]=min;
		}
		for(int i=0;i<m;i++){
			int x = min_c[i];
			while(min_c[i]>0){
				arli.add("col "+(i+1));
				min_c[i]--;
			}
			for(int j=0;j<n;j++)
				g[j][i]-=x;
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				if(g[i][j]!=0){
					System.out.println(-1);
					System.exit(0);
				}
			}
		}
		ArrayList<String> arli1 = new ArrayList<>();
		for(int i=0;i<m;i++){
			int min=Integer.MAX_VALUE;
			for(int j=0;j<n;j++){
				if(h[j][i]<min)
					min=h[j][i];
			}
			min_c[i]=min;
		}
		for(int i=0;i<m;i++){
			int x = min_c[i];
			while(min_c[i]>0){
				arli1.add("col "+(i+1));
				min_c[i]--;
			}
			for(int j=0;j<n;j++)
				h[j][i]-=x;
		}
		for(int i=0;i<n;i++){
			int min=Integer.MAX_VALUE;
			for(int j=0;j<m;j++){
				if(h[i][j]<min)
					min=h[i][j];
			}
			min_r[i]=min;
		}
		for(int i=0;i<n;i++){
			int x=min_r[i];
			while(min_r[i]>0){
				arli1.add("row "+(i+1));
				min_r[i]--;
			}
			for(int j=0;j<m;j++)
				h[i][j]-=x;
		}

		if(arli.size()<arli1.size()){
		System.out.println(arli.size());
		for(int i=0;i<arli.size();i++)
			System.out.println(arli.get(i));}
		else{
			System.out.println(arli1.size());
			for(int i=0;i<arli1.size();i++)
				System.out.println(arli1.get(i));
		}
	}
}